The focal length of a thin converging lens in | vedclass

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(A) In air,the lens maker's formula is $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Given $f_a = 20 \ cm = 0.2 \ m$ and

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$\frac{5}{3}$

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(A) In air,the lens maker's formula is $\frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Given $f_a = 20 \ cm = 0.2 \ m$ and $\mu_g = 1.5$,we have $\frac{1}{0.2} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \Rightarrow 5 = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \Rightarrow \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 10 \ m^{-1}$.When immersed in a liquid of refractive index $\mu_l$,the focal length $f_l$ is given by $\frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given the power $P = -1 \ D$,the focal length is $f_l = \frac{1}{P} = -1 \ m = -100 \ cm$.Substituting the values: $\frac{1}{-1} = \left( \frac{1.5}{\mu_l} - 1 \right) (10)$.$-0.1 = \frac{1.5}{\mu_l} - 1 \Rightarrow \frac{1.5}{\mu_l} = 0.9$.$\mu_l = \frac{1.5}{0.9} = \frac{15}{9} = \frac{5}{3}$.

The focal length of a thin converging lens in air is $20 \ cm$. When the lens is immersed in a liquid,it behaves like a concave lens of power $1 \ D$. If the refractive index of the material of the lens is $1.5$,the refractive index of the liquid is

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